Voltmeter/Ammeter Put the smoke back

[Jacksonbrown]

I received yesterday a Hiking dds238-2 sw. I ordered it for a completely different reason. It's a energy meter so it's measure kWh and it is DIN rail mountable. It can also measure Volts, Amps, frequency and real power. It's made for billing purposes and I found that it is very accurate. It also got an open collector output which give 1600 short pulses for each kWh. Connect it to an Arduino, measure the time between pulses and the power is 2250 / (time between pulses)

This looks like the kind of thing I've been looking for.
Something with an output and all the better for DIN mount.

What does the current value mean? (the 5/10 bit)

5(65)A, 5(30)A, 10(40)A, 5(60)A or other as required

[Edwin Croissant]

What does the current value mean? (the 5/10 bit)

5(65)A, 5(30)A, 10(40)A, 5(60)A or other as required

5(65) means the Ib=5A and the Imax=65A

The IEC1036 standard specifies accuracy over a range of 5% Ib to IMAX [..]The basic current [Ib] is defined in IEC1036 (1996–09) section 3.5.1.1 as the value of current in accordance with which the relevant performance of a direct connection meter is fixed. I MAX is the maximum current at which accuracy is maintained.

I tested this meter some more today and found that it is very accurate. The only drawback I find is that the meter displays kWh when switched on and that the button must be pressed several times to get the power reading. Which is irrelevant when connected to an Arduino .

Most of these din-rail energy meters have an open collector output. Also the cheaper one's with only lcd or mechanical(stepper motor operated register) kWh readout.

[skow69]

Damn you, Edwin. You did that on purpose.

1600 pulses for a kWh,
so a continuous rate of 1kW would be
1600 pulses per hour,
1600/3600 seconds per hour =
0.444 pulses per second, or
2.25 seconds per pulse, so
1 watt continuous would output 1 pulse every 2250 seconds.
Hence power (in watts) = 2250/pulse period (in seconds).

Whew! Finally!
I'm sure there is an easier way to get there, but now that I have that little puzzle out of my head, I can go back to thinking about whatever the hell it was we were talking about.

[NZChris]

I have an analogue ammeter daisy chained with the controller today, and it agrees with the cheapo digital, (wired to the input side), within 0.1 amps.

I'll take that back.

It tells lies when the controller is dialed back. I now only use the analogue ammeter.

[skow69]

I think I got it.

I am running the exact same setup as Cranky. If the connection is moved to the input side of the SSR, will it not just always read my maximum voltage being fed and not the cut voltage actually going to the heating element? I already know I have 237 volts feeding in. If I don't know the voltage and amperage output going to the element how do I determine the wattage?

That is exactly where I started from, and why I raised a fuss in Edwin's first post about the magic meter.

Edwin wrote:
This meter measures a couple of thousand times a second the actual voltage and current value. It multiplies the numbers which each other and add the results together. After a second the result is the real power. So it only measure the power when there is current flowing through your element. I am working on a simple explanation but that is quite a challenge When I studied this subject it was not one of the easiest one's

The rest of this will be "Lectricity for Dummys" written by a dummy, so if this is inaccurate I hope Edwin will show up quick to correct it, cuz the one thing that is worse than no information is wrong information.

The rapid sampling seems to be just a matter of accuracy, or resolution. More samples equals more accurate. The regulating agencies require an energy meter, used by the power supplier to bill customers, to read the waveform accurately to the 10th harmonic. The scientific community has agreed to accept the accuracy of a sample rate that is double what is required. This turns out to be 2k samples per second at 50Hz, 2.4k at 60Hz.

The meter measures voltage and current separately and calculates the waveform of each. [Picture the two waves superimposed on an oscilloscope screen.] They are not the same. We ignore the amplitude. The frequency is identical, but one is ahead of the other. This is called phase shift. The measure of the difference between them is the phase angle. Capacitance causes voltage to lag behind current. Inductance causes current to lag behind voltage. In either case the phase angle is between 0 and 90o.

Apparent Power is VA, a simple calculation. The actual energy available for the user to put to work is the Real Power, represented by the area under the curves, which is considerably harder to quantify. [The area under the curve for a true sine wave is 0.707 of the peak, which is the RMS value.]

The phase angle (0) is calculated from the waveform data. Mathematically, the power factor is given by the cosine of the phase angle. The power factor is also defined as real power (P) divided by apparent power (S). So cos0=P/S, and BINGO the meter has everything necessary to tell us how much power is going out by reading the inputs only.

There are a couple of things that still bother me, but I will keep them to myself for now to avoid muddying the waters with nonsense.

Edwin?

[Or anyone else who really understands this shit, but no one else has stepped up so far.]